Reverse int within the 32-bit signed integer range: $[−2^{31}, 2^{31} − 1]$ OptimizedReverse int within...
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Reverse int within the 32-bit signed integer range: $[−2^{31}, 2^{31} − 1]$ Optimized
Reverse int within the 32-bit signed integer range: $[−2^{31}, 2^{31} − 1]$
$begingroup$
LeetCode Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.
#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>
class Solution
{
public:
int reverse(int i) {
if(i > INT_MAX || i < INT_MIN) {
return 0;
}
int sign = 1;
if(i < 0) {
sign = -1;
i = i*sign;
}
int reversed = 0;
int pop = 0;
while(i > 0) {
pop = i % 10;
reversed = reversed*10 + pop;
i /= 10;
}
std::cout << reversed << 'n';
return reversed*sign;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(1534236469) == 0);
assert(s.reverse(-2147483412) == -2143847412);
}
c++ mathematics integer
asked 3 mins ago
greggreg
38218
$endgroup$
add a comment |
$begingroup$
LeetCode Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.
#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>
class Solution
{
public:
int reverse(int i) {
if(i > INT_MAX || i < INT_MIN) {
return 0;
}
int sign = 1;
if(i < 0) {
sign = -1;
i = i*sign;
}
int reversed = 0;
int pop = 0;
while(i > 0) {
pop = i % 10;
reversed = reversed*10 + pop;
i /= 10;
}
std::cout << reversed << 'n';
return reversed*sign;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(1534236469) == 0);
assert(s.reverse(-2147483412) == -2143847412);
}
c++ mathematics integer
asked 3 mins ago
greggreg
38218
$endgroup$
add a comment |
$begingroup$
LeetCode Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.
#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>
class Solution
{
public:
int reverse(int i) {
if(i > INT_MAX || i < INT_MIN) {
return 0;
}
int sign = 1;
if(i < 0) {
sign = -1;
i = i*sign;
}
int reversed = 0;
int pop = 0;
while(i > 0) {
pop = i % 10;
reversed = reversed*10 + pop;
i /= 10;
}
std::cout << reversed << 'n';
return reversed*sign;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(1534236469) == 0);
assert(s.reverse(-2147483412) == -2143847412);
}
c++ mathematics integer
asked 3 mins ago
greggreg
38218
$endgroup$
LeetCode Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.
#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>
class Solution
{
public:
int reverse(int i) {
if(i > INT_MAX || i < INT_MIN) {
return 0;
}
int sign = 1;
if(i < 0) {
sign = -1;
i = i*sign;
}
int reversed = 0;
int pop = 0;
while(i > 0) {
pop = i % 10;
reversed = reversed*10 + pop;
i /= 10;
}
std::cout << reversed << 'n';
return reversed*sign;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(1534236469) == 0);
assert(s.reverse(-2147483412) == -2143847412);
}
c++ mathematics integer
c++ mathematics integer
asked 3 mins ago
greggreg
38218
asked 3 mins ago
greggreg
38218
asked 3 mins ago
greggreg
38218
asked 3 mins ago
greggreg
38218
asked 3 mins ago
greggreg
38218
38218
add a comment |
add a comment |
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